TheoremProved

Vieta's formulas

Statement

For a monic polynomial $x^n + a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ with roots $r_1,\dots,r_n$ (counted with multiplicity), the elementary symmetric sums of the roots equal the coefficients up to sign: $\sum_i r_i = -a_{n-1}$ , $\sum_{i<j} r_ir_j = a_{n-2}$ , …, $r_1r_2\cdots r_n=(-1)^n a_0$ .

Why is it true?

If you already know a polynomial's roots, you can rebuild the polynomial by multiplying out $(x-r_1)(x-r_2)\cdots(x-r_n)$ . Expanding that product mixes the roots together in every possible way — sums of one root, sums of products of two roots, and so on — and those mixtures are exactly the coefficients. So the coefficients are not arbitrary numbers: they are a compressed record of how the roots combine.

Proof sketch

Expand $(x-r_1)(x-r_2)\cdots(x-r_n)$ and compare coefficients with $x^n+a_{n-1}x^{n-1}+\cdots+a_0$ ; the coefficient of $x^{n-k}$ on the left is $(-1)^k e_k(r_1,\dots,r_n)$ , where $e_k$ is the $k$ -th elementary symmetric polynomial, which must equal $a_{n-k}$ .

Topics that use this theorem

Quadratic equations

Related theorems

Quadratic formula

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Victor J. Katz (2009). A History of Mathematics: An Introduction